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Intuition mismatch

  1. Mar 2, 2012 #1
    This is probably a trivial problem, but I can't get it straight. It's well known and easy enough to show mathematically that the change in enthalpy for the reversible isothermal expansion of an ideal gas is always zero, ΔH=0. But it's also true that the enthalpy change for a process at constant P is equal to the heat that enters (or leaves) the system, ΔH=qP. Fine.
    So if I take a cylinder with a piston and place it in a heat bath, and pull the piston out in a slow, reversible fashion, I can allow that the temperature remains constant and the process is isothermal; it follows that ΔH = 0. Um...but didn't heat cross the boundary of the system? In fact doesn't it HAVE to, to keep the temperature constant? The whole process was expansion against a constant Pressure, so ΔH should equal the heat that entered the cylinder, a quantity definitely not zero.
    It seems to me that with ΔT=0 the internal energy is not changing, so q=-w, and here w=P dV. Again, this is a non-zero quantity.
    How do I reconcile this?
    (Note: Please don't derive for me that ΔH=0 from the ideal gas law. I know, and I'm convinced. But that doesn't explain why the reasoning with the cylinder is wrong.)
  2. jcsd
  3. Mar 3, 2012 #2
    Short version: The heat bath cools down.

    You're splitting up the problem differently in each case. The "system" is either the cylinder OR the cylinder and the heat bath. If you expand the cylinder but bring heat in from outside, you have to account for the heat LOST outside before you get to a zero-sum.
  4. Mar 3, 2012 #3
    That sounds like it might be the right track to me. Can I get a slightly longer version? I'm taking the system to be just the gas, which I am free to do, and the above reasoning still applies. It sounds like you're saying the solution is that ΔH=0 for an isothermal expansion ONLY if the system is isolated, not merely closed. Is this correct? I've been over the several sources that derive this statement and none of them come right out and say that. Am I right in gleaning that for a closed but not isolated system, ΔH=∫CPdT, and not zero? This is the only thing that makes intuitive sense to me.
    But wait. If the system is isolated, that makes it an adiabatic expansion, not an isothermal one. So which one does ΔH=0 apply to?
    This is such a simple physical problem, I'm surprised no one has been able to connect the pieces to the theory in a straightforward way.
    Last edited: Mar 3, 2012
  5. Mar 3, 2012 #4

    Philip Wood

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    The paradox, if I've got it right (and I probably haven't), is dependent on the pressure being constant:

    "But it's also true that the enthalpy change for a process at constant P is equal to the heat that enters (or leaves) the system, ΔH=qP."

    But I can't see how the pressure CAN be constant for the isothermal expansion you describe. Surely, at constant temperature, if V goes up, p must go down?
  6. Mar 3, 2012 #5
    Okay, I think the issue is that I'm imagining the expansion is against a constant pressure, which is perfectly reasonably physically, but the pressure in the gas will not be constant. So I can't take the enthalpy to be equal to the heat flow. Simply, ΔH ≠qP in this situation, and ΔH = 0 even though the there is heat flow across the cylinder.
    Is that right?
  7. Mar 4, 2012 #6

    Philip Wood

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    That's my reading of it. ΔT = 0 and Δ(PV) = 0, so ΔH = 0.
  8. Mar 5, 2012 #7
    Okay that makes sense. Thanks.
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